Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int arr[123][123];
int tmp[123][123];
int y, m, n, k = 0, k2 = 0;
struct L {
  int first, second, third;
};
vector<L> ans;
vector<L> ans2;
void col(int h) {
  for (int j = 0; j < m; j++) {
    int mn = 5000;
    for (int i = 0; i < n; i++) mn = min(arr[i][j], mn);
    if (h == 1)
      k += mn;
    else
      k2 += mn;
    L x;
    if (mn) {
      x.first = j + 1, x.second = mn, x.third = 2;
      if (h == 1)
        ans.push_back(x);
      else
        ans2.push_back(x);
    }
    for (int i = 0; i < n; i++) arr[i][j] -= mn;
  }
}
void row(int h) {
  for (int i = 0; i < n; i++) {
    int mn = 5000;
    for (int j = 0; j < m; j++) mn = min(arr[i][j], mn);
    if (h == 1)
      k += mn;
    else
      k2 += mn;
    if (mn) {
      L x;
      x.first = i + 1, x.second = mn, x.third = 1;
      if (h == 1)
        ans.push_back(x);
      else
        ans2.push_back(x);
    }
    for (int j = 0; j < m; j++) arr[i][j] -= mn;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> arr[i][j];
      tmp[i][j] = arr[i][j];
    }
  int y = m, x = n;
  row(1);
  col(1);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      arr[i][j] = tmp[i][j];
    }
  col(2);
  row(2);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (arr[i][j]) {
        cout << "-1" << endl;
        return 0;
      }
  if (k <= k2) {
    cout << k << endl;
    for (int i = 0; i < ans.size(); i++) {
      if (ans[i].third == 2) {
        y--;
        for (int j = 0; j < ans[i].second; j++)
          cout << "col " << ans[i].first << endl;
      } else {
        for (int j = 0; j < ans[i].second; j++)
          cout << "row " << ans[i].first << endl;
      }
    }
  } else {
    cout << k2 << endl;
    for (int i = 0; i < ans2.size(); i++) {
      if (ans2[i].third == 2) {
        for (int j = 0; j < ans2[i].second; j++)
          cout << "col " << ans2[i].first << endl;
      } else {
        for (int j = 0; j < ans2[i].second; j++)
          cout << "row " << ans2[i].first << endl;
      }
    }
  }
  return 0;
}